Question

Consider the circle centered at the origin and passing through the point (0, 2).

(a) Give the equation of the circle.

(b) For each point, decide whether or not it is on the circle. Is the point on the circle? (look at the image I attached)​

Answer 1

Answer:

x^2+y^2=4, yes, no, yes, no

Step-by-step explanation:

a) It's centered at the origin, which means that the equation is x^2+y^2=r^2, right? Then since point (0,2) is on the circle, its radius can be solved using Pythagoras theory, r=sqrt(0^2+2^2)=2, so the equation is x^2+y^2=4.

b) You plug the points into the equation and if the equation is true, they're on the circle. I'll give you an example of the first two.

($\sqrt{3},1$): so x=$\sqrt{3}$, y=1. When you bring that into the equation you got in (a), it would be 3+1=4, which is true, which means that the point is on the circle.

(4,0): so x=4, y=0. When you bring that into the equation you got in (a), it would be 16+0=4, which is not true, which means the point isn't on the circle.